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4 years ago

Plzz help me and I’ll help too!

Mathematics

1 answer:

dimaraw [331]4 years ago

6 0

Answer:

2; 2; 0; 1; Isosceles and Obtuse

Step-by-step explanation:

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Is Zero a Positive Number or a Negative Number or a Neutral Number?

sergeinik [125]

Its a neutral number cause it has no value since its worth nothing at all, when u add it with something the number ur adding it with will be that number instead of 0, (like for ex 2 + 0 = 2) 0 cant be positive since on a number line its not on the right and it cant be negative since its not on the left, so its neutral in the middle.

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3 years ago

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BEST ANSWER GETS BRAINLIEST!!

777dan777 [17]

2 over 3 plus 5 equals 5 and 2 over 3

3 0

3 years ago

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I need to know how to do it in Distributive Property.

kotykmax [81]

A(b-c)=ab-ac

(1/7)(1/3h-1/2)=(1/7)(1/3h)-(1/7)(1/2)=1/21h-1/14

now we gots
6 and 1/7+1/21h-1/14
rearange
6 and 1/7-1/14+1/21h
make common denom which is 14
6 and 2/14-1/14+1/21h
6 and 1/14+1/21h

6 0

3 years ago

List the steps that should be used to solve the equation f(x) = g(x) by graphing when f(x) = (1/2)² - 3 and g(x) = 3x + 5?

jolli1 [7]

value of x is: $x=\frac{-31}{12}$

Step-by-step explanation:

We need to solve the equation f(x)=g(x)

where $f(x)=(\frac{1}{2})^2-3$

and $g(x) = 3x + 5$

Solving to find f(x)=g(x)

Putting values of f(x) and g(x)

$f(x)=g(x)\\(\frac{1}{2})^2-3=3x + 5\\\frac{1}{4}-3=3x + 5\\Subtarct\,\,5\,\,on\,\,both\,\,sides:\\\frac{1}{4}-3-5=3x + 5-5\\\frac{1}{4}-8=3x\\Taking\,\,LCm\,\,on\,\,left\,\,side\\\frac{1-8*4}{4}=3x\\\frac{-31}{4}=3x\\Divide\,\,both\,\,sides\,\,by\,\,3\\x=\frac{-31}{4*3}\\x=\frac{-31}{12}$

So, value of x is: $x=\frac{-31}{12}$

The graph will be a straight line x= -2.58 or x = -31/12

Keywords: Composition of Functions

Learn more about Composition of Functions at:

#learnwithBrainly

7 0

3 years ago

Perform the indicated operation and express the result as a simplified complex number. I need help with 35

skelet666 [1.2K]

Given:-

$\frac{3+4i}{2-i}$

To find:-

The simplified form.

At first we take conjucate and multiply below and above.

The conjucate is,

$2+i$

So now we multiply. we get,

$\frac{3+4i}{2-i}\times\frac{2+i}{2+i}$

Now we simplify. so we get,

$\frac{3+4i}{2-i}\times\frac{2+i}{2+i}=\frac{6+3i+8i+4(i)^2}{2^2-i^2}$

We know the value of,

$i^2=-1$

Substituting the value -1. we get,

$\begin{gathered} \frac{6+3i+8i+4(i)^2}{2^2-i^2}=\frac{6+3i+8i+4(-1)_{}^{}}{2^2-(-1)^{}} \\ \text{ =}\frac{6-4+11i}{2+1} \\ \text{ =}\frac{2+11i}{3} \end{gathered}$

So now we split the term to bring it into the form a+ib. so we get,

$\frac{2+11i}{3}=\frac{2}{3}+i\frac{11}{3}$

So the required solution is,

$\frac{2}{3}+i\frac{11}{3}$

3 0

1 year ago